Ex. Find the 10th term of the A. P. whose 7th term is 15 and whose 21st term is 22. If a be the first term, and d be the common difference, we have 1 a+6d=15, and a +20d=22. 9 Hence d= a=12. The 10th term is therefore 12+ = 16. 218. When three quantities are in arithmetical progression, the middle one is called the Arithmetic Mean of the other two. If a, b, c are in A. P., we have, by definition, b-a=c-b; and therefore b = {(a+c). Thus the arithmetic mean of two given quantities is half their sum. When any number of quantities are in arithmetical progression all the intermediate terms may be called arithmetic means of the two extreme terms. Between any two given quantities any number of arithmetic means may be inserted. Let a and b be the two given quantities, and let n be the number of terms to be inserted. Then b will be the n + 2th term of the A. P. whose first term is a. Hence, if d be the common difference, b = a + (n + 1) d; b-a and therefore d= n+1' 219. To find the sum of any number of terms of an arithmetical progression. Let a be the first term and d the common difference. Let n be the number of the terms whose sum is required, and let 7 be the last of them. Then, since is the nth term, we have l = a + (n − 1) d................ Hence, if S be the required sum, ..(i). S= a + (a + d) + (a + 2d) +...... + (l − 2d) + (1 − d) + l. Now write the series in the reverse order; then S = l + (l − d) + (l − 2d) +........... + (a + 2d) + (a + d) + a. Hence, by addition of corresponding terms, we have 28=(a+1)+(a + 1) + (a + 1) + ...... to n terms = n(a + 1); From the formulæ (i), (ii), (iii) the value of all the quantities a, d, n, l, S can be found when any three are given. Ex. 1. Find the sum of 20 terms of the arithmetical progression 3+6+9+&c. Ex. 2. Shew that the sum of any number of consecutive odd numbers, beginning with unity, is a square number. The series of odd numbers is 1+ 3 + 5+...... Here a= =1, d=2; hence the sum of n terms is given by Ex. 3. How many terms of the series 1 + 5 + 9 + in order that the sum may be 190? must be taken {2a + (n − 1) d}, where S=190, a=1, d=4. Hence n is to be found from the quadratic equation 19 2 Hence n = 10. The value n = - is to be rejected for n must necessarily be a positive integer*. Ex. 4. How many terms of the series 5+7+9+...... must be taken in order that the sum may be 480. Hence n must be 20, for the value n = 24 must be rejected as a negative number of terms is altogether meaningless *. Ex. 5. What is the 14th term of the A. P. whose 5th term is 11 and whose 9th term is 7? Ans. 2. Ex. 6. What is the 2nd term of the A. P. whose 4th term is and whose 7th term is 3a + 4b? Ans. 2a-b. Ex. 7. Which term of the series 5, 8, 11, &c. is 320? Ans. The 106th. Ex. 8. Shew that, if the same quantity be added to every term of an A. P., the sums will be in A. p. Ex. 9. Shew that, if every term of an A. P. be multiplied by the same quantity, the products will be in A. P. * The inadmissible value is a root of the equation to which the problem leads, but it is not a solution of the problem. [See Chapter x1.] It should be remarked that a negative value of n cannot mean a number of terms reckoned backwards. Ex. 10. Shew that, if between every two consecutive terms of an A. P., a fixed number of arithmetic means be inserted, the whole will form an arithmetical progression. Ans. (i) 621, (ii) – 16, (iii) 10a, (iv) — (n−1). Ex. 12. The 7th term of an A. P. is 15, and the 21st term is 8; find Ans. 420. Ex. 14. Shew that, if any odd number of quantities are in a. p., the first, the middle and the last are in A. P. Ex. 15. Shew that, if unity be added to the sum of any number of terms of the series 8, 16, 24, &c., the result will be the square of an odd number. Ex. 16. How many terms of the series 15 + 11 + 7 + taken in order that the sum may be 35? Ex. 17. The sum of 5 terms of an A. P. is - 13; what is the common difference? must be Ans. 5. Ex. 18. Find the sum of all the numbers between 200 and 400 which are divisible by 7. Ans. 8729. Ex. 19. If a series of terms in A. P. be collected into groups of n terms, and the terms in each group be added together, the results form an A. P. whose common difference: the original common difference as n2: 1. GEOMETRICAL PROGRESSION. 220. Definition. A series of quantities is said to be in Geometrical Progression when the ratio of any term to the preceding one is the same throughout the series. Thus a, b, c, d, &c. are in Geometrical Progression The ratio of each term of a geometrical progression to the preceding term is called the common ratio. The following are examples of geometrical progressions: In the first series the common ratio is 3, in the second series it is - 1⁄2, and in the third series it is a2. 221. If the first term of a G. P. be a, and the common ratio r; then, by definition, and so on, the index of r being always less by unity than the number giving the position of the term in the series. Hence the nth term will be ar2-1. We can therefore write down any term of a G. P. when the first term and the common ratio are given. For example, in the G. P. whose first term is 2, and whose common ratio is 3, the 6th term is 2 × 35, and the 20th term is 2 × 319. 222. A Geometrical Progression is determined when any two of its terms are given. For, suppose we know that the mth term is a, and that the nth term is ß. Let a be the first term, and r the common ratio; then the mth term will be arm-1, and the nth term will be arn-1. |